Hello!
We know that this is a right triangle and the angle C is 45º.
Knowing it, we have:
Considering the information above, we must use the sine of 45º to calculate the value of side BC, look:
[tex]\sin(45\degree)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}[/tex]
As we know, the sine of 45º is:
[tex]\sin(45)=\frac{\sqrt{2}}{2}[/tex]
Let's replace all the values in the formula:
[tex]\begin{gathered} \sin(45\operatorname{\degree})=\frac{\mathrm{oppos\imaginaryI te}}{\mathrm{hypotenuse}} \\ \\ \dfrac{\sqrt{2}}{2}=\frac{10}{\mathrm{BC}} \\ \\ \mathrm{BC}\sqrt{2}=10\cdot2 \\ \mathrm{BC}\sqrt{2}=20 \\ BC=\frac{20}{\sqrt{2}} \\ \\ BC=\frac{20\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}}=\frac{20\sqrt{2}}{\sqrt{4}}=\frac{20\sqrt{2}}{2}=\boxed{10\sqrt{2}\text{ ft}} \end{gathered}[/tex]
Answer:
Alternative B.
